Find the area of the smaller part of the circle `x^2+y^2=a^2`cut off by the line `x=a/(sqrt(2))`

Options

A. `(a)/(2)((pi)/(2)+1)` sq units
B. `(a^(2))/(2)((pi)/(2)-1)` sq units
C. `a((pi)/(2)-1)` sq units
D. None of these

Answer

B

Explanation

Given line, `x=(a)/(sqrt2)" ... (i)"`
and circle, `x^(2)+y^(2)=a^(2)" ... (ii)"`
Since, given line cuts the circle.
So, put `x=(a)/(sqrt(2))` in Eq. (ii), we get

`((a)/(sqrt(2)))^(2)+y^(2)=a^(2)impliesy^(2)=(a^(2))/(1)-(a^(2))/(2)=(a^(2))/(2)`
`implies |y|=(a)/(sqrt(2))impliesy=pm(a)/(sqrt(2))`
`therefore` Intersection point in first quadrant is `((a)/(sqrt(2)),(a)/(sqrt(2)))`.
Required area
= 2 (Area of shaded region in first quadrant only)
`=2int_(a//sqrt(2))^(a)|y|dx=2int_(a//sqrt(2))^(a)sqrt(a^(2)-x^(2))dx`
`=2[(x)/(2)sqrt(a^(2)-x^(2))+(a^(2))/(2)sin^(-1)((x)/(a))]_(a//sqrt(2))^(a)`
`=2[0+(a^(2))/(2)sin^(-1)(1)-(a)/(2sqrt(2))sqrt(a^(2)-(a^(2))/(2))-(a^(2))/(2)sin^(-1)((1)/(sqrt(2)))]`
`=2[(a^(2))/(2)((pi)/(2))-(a)/(2sqrt(2)).(a)/(sqrt(2))-(a^(2))/(2)((pi)/(4))]`
`=2[(a^(2)pi)/(4)-(pia^(2))/(8)-(a^(2))/(4)]=((a^(2)pi)/(4)-(a^(2))/(2))=(a^(2))/(2)((pi)/(2)-1)` sq units